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PHYS 414 Final Exam
Problem 1/3: Quantum Jarzynski Equality
Earlier this year, researchers used a single ion trapped in a harmonic potential to verify for
the first time a quantum version of the Jarzynski equality [An et al., Nature Physics 11, 193
(2015); http://arxiv.org/abs/1409.4485]. Beyond the experimental challenges of manipulating ions
at micro-Kelvin temperatures, there is a fundamental theoretical puzzle: extending non-equilibrium
results like the Jarzynski equality to the quantum regime is non-trivial. In open quantum systems which interact with the environment, there is an ongoing debate on how to properly define
thermodynamic concepts like work and heat in the language of quantum mechanics (work is not
an observable!) There is however one special case where a quantum Jarzynski equality can be
unambiguously derived, which we will explore in this problem.
ˆ controlled by the experConsider a quantum ensemble with a time-dependent Hamiltonian H(t)
imentalist. There are several steps in the experiment:
ˆ
ˆ init for all t ≤ 0. The system is
1. For t ≤ 0 the Hamiltonian is kept constant: H(t)
= H
coupled to the environment and allowed to reach equilibrium at some temperature T . At
t = 0 the density operator ρˆ in the energy basis is:
X
ρˆ(0) =
pinit
n |nihn|
n
ˆ init with eigenvalues En , and pinit = e−βEn /Zinit , with
where |ni
the eigenstates of H
n
P are
Zinit = n e−βEn and β = 1/kB T .
2. At t = 0 the experimentalist measures the energy of the system, and obtains one of the
eigenvalues En .
3. Between 0 < t < τ the experimentalist decouples the system from the environment, so
there is no longer the possibility of energy exchange or dephasing interactions. An external
ˆ
control parameter is varied, making H(t)
change with time. The overall time evolution of
this closed quantum system is described by a unitary time operator Uˆ (t), so that |ψ(t)i =
Uˆ (t)|ψ(0)i for any quantum state |ψ(0)i of the system at t = 0.
4. At time t = τ the experimentalist again measures the energy of the system. The Hamiltoˆ )=H
ˆ fin , with a different set of eigenstates and eigenvalues than H
ˆ init . Let
nian is now H(τ
us denote these by |νi and Eν , so the result of this measurement is some energy Eν .
a) What is the probability P (n) of measuring the value En in step 2?
b) What is the density matrix ρˆ(τ ) at time t = τ , right before the second measurement?
c) What is the probability P (ν)P
of measuring the value Eν in step 4? Check that your answer
for P (ν) has the form P (ν) = n P (ν|n)P (n), where P (ν|n) is the conditional probability of
measuring Eν at time t = τ , given that you measured En at time t = 0. Identify the expression
for P (ν|n).
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d) We can define the work W done on the system as the difference in energy between the two
measurements, W = Eν −En , since there is no heat Q exchanged with the environment between
t = 0 and t = τ . Calculate the average he−βW i, defined as:
X
he−βW i =
e−β(Eν −En ) P (ν|n)P (n)
ν,n
Show that it yields the Jarzynski result:
he−βW i = e−β∆F
s
s
s
where ∆F = Ffin
− Finit
. Here Finit
= −kB T ln
is the initial equilibrium free energy of
PZinit−βE
s
ν
the system, and Ffin = −kB T ln Zfin = −kB T ln ν e
. The latter is the free energy which
ˆ fin for t > τ , and the system was
the system would achieve if the Hamiltonian was fixed at H
recoupled to the environment so it equilibrates again at temperature T .
e) Show that the result of part d implies a quantum version of the second law of thermodynamics,
hW i ≥ ∆F . Hint: Would you like a Danish inequality to go with your coffee?
f) If you have gotten this far, you have seen how the proof of the quantum Jarzynski equality
is surprisingly elegant (we did not even have to invoke detailed balance!) This version of the
equality (with the system isolated from the environment in step 3) is what was experimentally
verified in the paper mentioned earlier. What if we became more ambitious and tried to find a
more general quantum Jarzynski equality, where we did not isolate the system in step 3? In this
case we know that instead of unitary time evolution, we can describe the change in the density
operator between t = 0 and t = τ as a general superoperator in Kraus form,
X
ˆ k ρˆ(0)M
ˆ†
M
ρˆ(τ ) =
k
k
ˆ †M
ˆ k are Kraus operators that satisfy P M
ˆ k = I,
ˆ and Iˆ is the identity operator. Try
where M
k
k
to redo parts c and d, and show that in general you unfortunately no longer get hexp(−βW )i =
exp(−β∆F ).
So any generalization of the Jarzynski equality to open quantum systems requires more work,
and this remains an active research area. For a summary of recent efforts, and one (of many)
contributions to this search, see http://arxiv.org/abs/1112.1303.
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Problem 2/3: An informed demon is an efficient demon
One of the most famous thought experiments involving a Maxwell demon was
concocted by Leó Szilárd in 1929 (the
same Szilárd who ten years later would
ghost-write Einstein’s letter to Franklin
Roosevelt, the genesis of the Manhattan Project). The so-called Szilard engine (Fig. 1) converts heat from a thermal reservoir into work, with the aid of
Figure 1: The Szilard engine model of a Maxwell
a “very observant and neat-fingered bedemon. Image credit: http://www.theo2.physik.uniing.” This model will allow us to explore
stuttgart.de/forschung/stochasticthermodynamics/
the role of “feedback” in controlling thermodynamic systems. The cycle of the engine is as follows:
1. At t = 0 we have a one-particle classical ideal gas in a box of volume V0 , at thermal equilibrium with temperature T .
2. The demon quickly inserts a thin partition in the middle of the box, splitting it into two
equal volumes of V0 /2. It waits until the gas equilibrates again.
3. At t = tM the demon makes a measurement of the particle position, figuring out whether
it is on the left (L) or right (R) side of the partition. For simplicity, we will initially assume
this measurement is perfect, and the demon never makes an error.
4. If the measurement was R, the demon allows the partition to very slowly slide to the left
between time t = tM and tS , such that the gas remains instantaneously in equilibrium as
the volume V increases from V0 /2 to V0 . If the measurement was L, the demon does the
same, but lets the partition slide slowly to the right. This step constitutes feedback control,
since the behavior of the demon depends on the outcome of the measurement.
5. At time t = tS the partition is removed, and the system returns to its initial equilibrium in
Step 1, which is achieved by some time t = τ .
The whole process between t = 0 and τ constitutes one full cycle of the engine.
a) If sliding the partition in and out of the box can be done without expending energy (similar to
the frictionless trap door in Maxwell’s original formulation) then the work in the engine consists
entirely of the expansion process described in Step 4. Answer the following questions: i) What is
the mean energy E¯ of the gas before and after the expansion? Call the difference the Rnet change
¯ S ) − E(t
¯ M ). ii) What is the total work done on the system W = tS dtW
˙ (t)
in energy ∆E¯ = E(t
tM
during the expansion? You should find that W < 0, soRactually the system does −W > 0
t
˙
work on its environment. iii) What is the total heat Q = tMS dtQ(t)
that flows into the system
during this process? You should find Q > 0. iv) What is the change in Helmholtz
free energy
Rt
∆F = F (tS ) − F (tM )? v) What is the internal entropy production S i = tMS dtS˙ i ? Hint: Dust
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˙ (t) = −f¯(t)x(t),
off the ideal gas law. Remember in our formalism W
˙
where x(t) is the control
¯
variable and f (t) is the mean conjugate force. In this case x = V and f = P . Do not forget that
the expansion is happening very slowly.
The answer to part a seems to violate the Kelvin-Planck statement of the second law, because the
engine operates in a cycle and does net work on its environment (W < 0) with no other effect than
drawing heat from a reservoir (Q > 0). However we know from Problem Set 5 that any realistic
demon must store the measured information somewhere (the demonic mind, a tape register of
bits, etc.), and ultimately erase it, leading to an eventual heat dump into the environment, and an
increase in the entropy of the universe. But exactly how much information is the demon storing?
Does the extracted work depend on the information content?
b) To get a better grasp on these issues, imagine now that the demon’s measurement apparatus
is imperfect. Let M denote the value which the demon gets in the measuring device (M = L or
R), and x denote the actual position of the particle (x = L or R). The conditional probability
P (M = L|x = L) = p, where 1/2 ≤ p ≤ 1, is the chance of getting a value of L in the measuring
device given that the particle is actually in L. Similarly P (M = R|x = R) = p. If p = 1 the device
is perfect, and we have the results of part a. If p < 1 the demon occasionally makes a mistake,
and slides the partition such that the volume of the gas is contracting instead of expanding. The
mutual information I which the demon obtains from the measurement, and which it can use for
feedback control, is given by:
I = DKL (P (M, x)||P (M )P (x)) =
R
X
M,x=L
P (M, x) ln
P (M, x)
P (M )P (x)
P
Here P (M, x) = P (M |x)P (x) is the joint
probability
of
M
and
x,
P
(M
)
=
x P (M, x) is the
P
marginal probability of M , and P (x) = M P (M, x) is the marginal probability of x. Find an
expression for I in terms of p. Check that when p = 1/2, the case where there is no correlation
between the measurement outcome and position, then P (M, x) = P (M )P (x) and the demon
gets zero mutual information. On the other extreme, check that when p = 1, the mutual information is largest. Note: Since we used ln instead of log2 in the definition of I, it is measured
in nats, instead of bits. Nats are the “natural” units when using information in the context of
thermodynamics. To convert to bits, you would divide I by ln 2.
c) To compensate for the error-prone measurement apparatus, let us give the demon a little flexibility in its feedback protocol (we will see why this is important shortly). We modify Step 4 above
as follows: if the measurement is R, the demon slides the partition to the left until the volumes of
the two regions in the box are V0 − (on the right) and (on the left), where 0 ≤ ≤ V0 /2 is some
fixed constant. Similarly if the measurement is L, the demon slides the partition to the right until
the volumes of the two regions are V0 − (on the left), and (on the right). The original protocol
corresponds to = 0. Find the average work W done on the gas, which will now be a function of
. Show that this attains a minimum (becoming maximally negative) when = V0 (1 − p). Check
that the value at this optimum is −W = kB T I, where I is the mutual information calculated in
part b.
d) What would happen if instead of allowing the partition to slide very slowly, the demon can
slide it at any speed it chooses, so the gas no longer has to remain approximately in equilibrium
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at all times during the expansion (or compression). Argue that this changes the result of part c
into an inequality: −W ≤ kB T I. Thus for best performance, the demon should keep the sliding
of the partition slow.
Remarkably, the demon can seemingly operate in flagrant violation of Kelvin-Planck (at least
while it has available space in its memory), so long as the demon can extract some nonzero
mutual information I > 0 about the position of the particle. To maximize the efficiency of the
engine, it has to choose a particular optimum feedback protocol, which depends on how accurate
the measurements are. Though we derived the results for the specific case of a Szilard engine,
you can actually establish a generalized second law for systems with feedback control: it states
that −W ≤ kB T I − ∆Ftot where ∆Ftot = F (τ ) − F (0) is the overall free energy difference of the
entire process. (For a cycle, like the Szilard engine, ∆Ftot = 0.) This version of the second law has
been beautifully demonstrated in an experimental system involving real-time feedback control
of a Brownian particle [Toyabe et al., Nature Physics 6, 988 (2010)]. Of course, as mentioned
above, we cannot indefinitely evade the original second law: eventually whatever information
the demon gathers will need to be erased, leading to heat being deposited into the environment.
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Problem 3/3: Einstein coefficients
One of the most elegant uses of detailed balance in physics was Einstein’s 1917 argument that
atoms must have two modes of emitting photons: spontaneous and stimulated emission. In this
problem we will recreate his proof.
a) Consider a cubic volume V = L3 filled with a gas of photons. The walls of this volume can
absorb and emit photons, so the gas equilibrates at some temperature T . Show that the mean
number of photons with energy E in equilibrium can be written as:
n
¯ s (E) =
g(E)
,
−1
eβE
where g(E) is the density of states. Find g(E). Hint: As we discussed in the review, photons are
bosons with energy E = ~c|k|, and in this photon gas scenario there is no number conservation,
so we have no chemical potential µ. Remember that for each possible wavevector k there are two
photon states (two possible polarizations).
b) Now let us focus on an individual two-level atom on the wall, equilibrated at temperature
T through interactions with the photon gas. (For simplicity we will neglect interactions with
other atoms on the wall.) Denote the ground and excited energies of the atom as E0 and E1 ,
with E1 − E0 = ~ω > 0. Let us assume that the two energy levels are possibly degenerate, as
is often the case: there are γ0 states with energy E0 , and γ1 states with energy E1 . Let W10 be
the effective transition rate between the entire group of states with energy E0 to the group of
states with energy E1 , and W01 be the effective transition rate in the opposite direction. Einstein
assumed that the rate W10 depends on the absorption of a photon of energy ~ω, driving the atom
from E0 to E1 . Hence W10 should be proportional to the number of photons with energy ~ω, and
we can write W10 = B10 n
¯ s (~ω), where B10 is a constant of proportionality. Use detailed balance
to show that this reasonable assumption implies a necessary form for the other transition rate,
W01 = A01 + B01 n
¯ s (~ω)
for some other constants A01 and B01 . How are these constants related to B10 ? Hint: To use
detailed balance correctly in the presence of degeneracy, figure out what is the equilibrium probability ps (E0 ) of observing any state with energy E0 , and analogously find ps (E1 ).
Collectively the constants A01 , B01 , and B10 are known as the Einstein coefficients. The A01 term
in W01 is not proportional to n
¯ s (~ω), so it constitutes a rate of spontaneous emission. The B01 term
in W01 is proportional to n
¯ s (~ω), so this constitutes stimulated emission: the atom is induced to
jump down in energy, releasing a photon, by the presence of other photons with energy ~ω. The
latter process is crucial to the mechanism of the laser.
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